7.4 Oxides of Nitrogen Nitrogen forms a number of oxides in different oxidation states. The names, formulas, preparation and physical appearance of these oxides are given in Table 7.3. Table 7.3: Oxides of Nitrogen (Name; Formula; Oxidation state of nitrogen; Common methods of preparation; Physical appearance and chemical nature) Dinitrogen oxide [Nitrogen(I) oxide]; N2O; + 1; NH4NO3 →(Heat) N2O + 2H2O; colourless gas, neutral Nitrogen monoxide [Nitrogen(II) oxide]; NO; + 2; 2NaNO2 + 2FeSO4 + 3H2SO4 → Fe2(SO4)3 + 2NaHSO4 + 2H2O + 2NO; colourless gas, neutral Dinitrogen trioxide [Nitrogen(III) oxide]; N2O3; + 3; 2NO + N2O4 →(250 K) 2N2O3; blue solid, acidic Nitrogen dioxide [Nitrogen(IV) oxide]; NO2; + 4; 2Pb(NO3)2 →(673K) 4NO2 + 2PbO + O2; brown gas, acidic Dinitrogen tetroxide [Nitrogen(IV) oxide]; N2O4; + 4; 2NO2 ⇌(Cool / Heat) N2O4; colourless solid/ liquid, acidic Dinitrogen pentoxide [Nitrogen(V) oxide]; N2O5; +5; 4HNO3 + P4O10 → 4HPO3 + 2N2O5; colourless solid, acidic Lewis dot main resonance structures and bond parameters of oxides are given in Table 7.4. Table 7.4: Structures of Oxides of Nitrogen (bond parameters): N2O: N–N 113 pm, N–O 119 pm, Linear. NO: N–O 115 pm. N2O3: N–N 186 pm, N–O 114 pm and 121 pm, angles 105°, 117°, 130°, Planar. NO2: N–O 120 pm, angle 134°, Angular. N2O4: N–N 175 pm, N–O 121 pm, angle 135°, Planar. N2O5: N–O 151 pm and 119 pm, angles 112° and 134°, Planar.
-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 177Phenol Electrophilic Substitution
Phenol Electrophilic Substitution, explained for NEET
The –OH group on a benzene ring activates the ring powerfully toward electrophilic substitution. This activation is the core of phenol reactivity in NEET, and the common confusion — treating –OH as just another activating group without accounting for the resonance-versus-induction balance in substituted phenols — costs marks reliably.
Why phenol is highly activated. The oxygen lone pair delocalises into the ring, increasing electron density at ortho and para positions. This makes phenol far more reactive than benzene toward electrophiles. NCERT Class 12 Chemistry Chapter 7 (Alcohols, Phenols and Ethers), page 206, documents this activation and the resulting ortho/para directing nature.
Bromination — the classic test. Phenol reacts with bromine water (Br₂/H₂O) at room temperature to give 2,4,6-tribromophenol (a white precipitate). No Lewis acid catalyst is needed — contrast this with benzene, which requires FeBr₃. This difference is a direct NEET question target. For mono-bromination, use Br₂ in CS₂ at low temperature, which gives predominantly para-bromophenol due to steric preference.
Nitration — temperature controls the product. Dilute HNO₃ at low temperature gives a mixture of ortho- and para-nitrophenol. Concentrated HNO₃ or a mixture of conc. HNO₃ + conc. H₂SO₄ gives 2,4,6-trinitrophenol (picric acid).
Kolbe's reaction (electrophilic carboxylation). Sodium phenoxide treated with CO₂ under pressure at 125°C gives sodium salicylate (ortho-hydroxybenzoic acid) after acidification. This is an electrophilic substitution unique to phenols.
Reimer-Tiemann reaction. Phenol + CHCl₃ + NaOH gives salicylaldehyde (ortho-hydroxybenzaldehyde) via an intermediate dichlorocarbene electrophile.
The substituted-phenol trap. When NEET asks you to predict the product of electrophilic substitution on a substituted phenol, you must consider both the –OH directing effect and the substituent's effect. Ignoring the resonance contribution and relying solely on inductive effects leads to wrong position predictions — this is a documented distractor pattern (observed in 2021, 2022, 2023, 2025 papers).
Can you answer these Phenol Electrophilic Substitution MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Phenol reacts with bromine water at room temperature to form:
Show answer and why every option is right or wrong
Answer: C. C is correct. Phenol is so strongly activated that aqueous bromine substitutes at all three available ortho/para positions without a catalyst, yielding 2,4,6-tribromophenol as a white precipitate (NCERT Class 12 Chemistry Chapter 7, page 211).
Why A is wrong: A is wrong — mono-bromination at the ortho position occurs with Br₂ in CS₂ (non-polar solvent, low temperature), not with aqueous bromine.
Why B is wrong: B is wrong — mono-bromination at para occurs preferentially with Br₂/CS₂; bromine water gives trisubstitution, not mono.
Why D is wrong: D is wrong — disubstitution is not the observed product with bromine water; the strong activation of the –OH group drives substitution at all three ortho/para positions.
Which catalyst is required for bromination of phenol with bromine water?
Show answer and why every option is right or wrong
Answer: D. D is correct. The –OH group activates the benzene ring so strongly through resonance donation that phenol reacts with bromine water at room temperature without any Lewis acid catalyst (NCERT Class 12 Chemistry Chapter 7, page 211).
Why A is wrong: A is wrong — AlCl₃ is a Friedel-Crafts catalyst used for benzene and deactivated arenes; phenol's high reactivity makes it unnecessary.
Why B is wrong: B is wrong — FeBr₃ is the standard Lewis acid catalyst for benzene bromination, but phenol does not require it.
Why C is wrong: C is wrong — ZnCl₂ is used in Lucas test for alcohols, not as a bromination catalyst for phenol.
The product of Kolbe's reaction (sodium phenoxide + CO₂ at 125°C under pressure, followed by acidification) is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Kolbe's reaction involves electrophilic carboxylation of sodium phenoxide at the ortho position, yielding sodium salicylate, which gives salicylic acid (2-hydroxybenzoic acid) upon acidification (NCERT Class 12 Chemistry Chapter 7).
Why A is wrong: A is wrong — phenyl benzoate is an ester formed by esterification, not by CO₂ carboxylation.
Why C is wrong: C is wrong — benzoic acid lacks the hydroxyl group; Kolbe's reaction preserves the –OH and introduces –COOH at the ortho position.
Why D is wrong: D is wrong — catechol has two –OH groups; Kolbe's reaction introduces a –COOH group, not a second –OH.
To obtain predominantly para-bromophenol from phenol, the appropriate reagent and solvent are:
Show answer and why every option is right or wrong
Answer: C. C is correct. In a non-polar solvent like CS₂ at low temperature, only monosubstitution occurs, and the para product predominates over ortho due to lower steric hindrance (NCERT Class 12 Chemistry Chapter 7, page 211).
Why A is wrong: A is wrong — bromine water is a polar medium that enables the high reactivity of phenol, leading to 2,4,6-tribromophenol, not monobromination.
Why B is wrong: B is wrong — NBS (N-bromosuccinimide) is used for allylic or benzylic bromination, not for electrophilic aromatic substitution on phenol.
Why D is wrong: D is wrong — FeBr₃ is a Lewis acid catalyst for less reactive arenes. For phenol, it is unnecessary and would not selectively give para-monobromination; the solvent choice (non-polar, low T) is what controls mono-selectivity.
Phenol is treated with CHCl₃ and NaOH, followed by acidification. The product is:
Show answer and why every option is right or wrong
Answer: A. A is correct. This is the Reimer-Tiemann reaction. CHCl₃ + NaOH generates dichlorocarbene (:CCl₂), which acts as the electrophile and substitutes at the ortho position of phenoxide. Hydrolysis of the intermediate gives salicylaldehyde (NCERT Class 12 Chemistry Chapter 7).
Why B is wrong: B is wrong — salicylic acid is the product of Kolbe's reaction (CO₂ + sodium phenoxide), not Reimer-Tiemann. Confusing these two named reactions is a common trap in NEET.
Why C is wrong: C is wrong — anisole is phenol's methyl ether (C₆H₅OCH₃), formed by Williamson synthesis with CH₃I/NaOH, not by CHCl₃/NaOH.
Why D is wrong: D is wrong — aspirin (acetylsalicylic acid) is made by acetylation of salicylic acid with acetic anhydride, not by Reimer-Tiemann reaction.
Among the following substituted phenols, which is the most acidic?
Show answer and why every option is right or wrong
Answer: A. A is correct. The –NO₂ group is a strong electron-withdrawing group that stabilises the phenoxide ion through both inductive and resonance withdrawal, making p-nitrophenol the most acidic among the options. The other substituents are electron-donating (–CH₃ by hyperconjugation, –OCH₃ by resonance), which destabilise the phenoxide ion and reduce acidity (NCERT Class 12 Chemistry Chapter 7).
Why B is wrong: B is wrong — the –CH₃ group is electron-donating (hyperconjugation/inductive), which destabilises phenoxide and makes p-cresol less acidic than phenol itself.
Why C is wrong: C is wrong — unsubstituted phenol is less acidic than p-nitrophenol because it lacks the electron-withdrawing stabilisation of phenoxide.
Why D is wrong: D is wrong — the –OCH₃ group is electron-donating by resonance (stronger than its weak –I effect), destabilising phenoxide and making p-methoxyphenol less acidic than phenol.
Phenol undergoes electrophilic substitution much more readily than benzene because:
Show answer and why every option is right or wrong
Answer: B. B is correct. The oxygen's lone pair enters resonance with the ring π-system, creating partial negative charge at ortho and para carbons. This increased electron density makes the ring more nucleophilic and more attractive to electrophiles (NCERT Class 12 Chemistry Chapter 7, page 211).
Why A is wrong: A is wrong — the reactivity of phenol toward electrophilic substitution relates to ring electron density, not O–H bond strength. Bond strength is relevant in acidity discussions, not in ring substitution.
Why C is wrong: C is wrong — molecular weight has no bearing on electrophilic substitution reactivity. Reactivity depends on the electronic effect of the substituent on the ring.
Why D is wrong: D is wrong — while –OH does have a weak inductive electron-withdrawing effect, its dominant effect is resonance donation INTO the ring, which activates it. Treating –OH as purely electron-withdrawing leads to incorrect predictions of deactivation.
Phenol is first treated with dilute HNO₃ at low temperature, and the para-nitrophenol product is then isolated. If this para-nitrophenol is now brominated with Br₂/H₂O, the expected product is:
Show answer and why every option is right or wrong
Answer: D. D is correct. In para-nitrophenol, the –OH at position 1 is ortho/para directing and strongly activating. The –NO₂ at position 4 is meta directing and deactivating. The positions ortho to –OH (positions 2 and 6) are also meta to –NO₂, making them doubly favoured. Position 4 is already occupied. Bromine water, being a strong enough system for activated phenols, substitutes at both positions 2 and 6.
Why A is wrong: A is wrong — with bromine water (excess Br₂ in aqueous medium), the phenol ring is sufficiently activated to undergo bromination at both available ortho positions (2 and 6), not just one. Monobromination would require controlled conditions (Br₂/CS₂, low T).
Why B is wrong: B is wrong — 2,4,6-tribromophenol is the product from unsubstituted phenol + Br₂/H₂O. Position 4 here is already blocked by –NO₂, so trisubstitution at 2,4,6 is impossible.
Why C is wrong: C is wrong — this places bromine at position 4, which is already occupied by –NO₂. Furthermore, bromine water with a phenol delivers substitution at the most activated positions, which are 2 and 6.
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How do you solve a Phenol Electrophilic Substitution question? A worked example
- 1
Given
Four substituted phenols: phenol (C₆H₅OH), p-nitrophenol (O₂N–C₆H₄–OH), p-cresol (CH₃–C₆H₄–OH), p-chlorophenol (Cl–C₆H₄–OH).
- 2
Required
Decreasing order of acidity.
- 3
Concept
Acidity of phenols depends on stability of the phenoxide ion formed after proton loss. Electron-withdrawing groups (EWG) stabilise phenoxide by dispersing negative charge → increase acidity. Electron-donating groups (EDG) destabilise phenoxide → decrease acidity. Both resonance and inductive effects must be considered; ignoring one leads to wrong ordering.
- 4
Principle applied
For para-substituents on phenol:• –NO₂ is strongly electron-withdrawing by both –I and –R effects → strongly stabilises phenoxide.• –Cl is electron-withdrawing by –I effect but weakly electron-donating by +R effect. Net effect at the para position: weak electron withdrawal → mildly stabilises phenoxide.• –CH₃ is electron-donating by hyperconjugation and +I effect → destabilises phenoxide.
- 5
Analysis (substituent by substituent)
• p-Nitrophenol: –NO₂ has strong –I and –R → most stable phenoxide → most acidic.• p-Chlorophenol: –Cl has –I (moderate) and weak +R → net mild stabilisation → more acidic than phenol.• Phenol: no substituent → reference point.• p-Cresol: –CH₃ has +I and hyperconjugation → destabilises phenoxide → least acidic.
- 6
Ordering
p-Nitrophenol > p-Chlorophenol > Phenol > p-Cresol
- 7
Final answer
Decreasing acidity: p-Nitrophenol > p-Chlorophenol > Phenol > p-Cresol
- 8
Common trap
The documented distractor pattern here is "ignores resonance-vs-induction balance." A student relying only on electronegativity might rank –Cl as strongly acidifying (comparable to –NO₂) or might forget that –CH₃ is electron-donating. The resonance effect of –NO₂ withdrawing through the ring is far stronger than the inductive effect of –Cl alone.
- 9
Similar NEET-style question
Arrange in increasing order of acidity: 2,4-dinitrophenol, m-cresol, phenol, p-fluorophenol. (Same principle — EWG/EDG effects on phenoxide stability — but with a disubstituted phenol and a halogen at a different position.)
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What to remember before solving Phenol Electrophilic Substitution questions
More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 3 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
Phenol Electrophilic Substitution questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 23 past-paper questions from Alcohols, Phenols, Ethers and Carbonyl Compounds →
How does NEET ask about Phenol Electrophilic Substitution?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Predict acidity ordering of substituted phenols, or product of nitration/halogenation.
Common distractors
ignores resonance vs induction balance
Uses inductive only without considering resonance
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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